Question: The side of the base of a square prism is increasing at a rate of $5$ meters per second and the height of the prism is decreasing at a rate of $2$ meters per second. At a certain instant, the base's side is $6$ meters and the height is $7$ meters. What is the rate of change of the volume of the prism at that instant (in cubic meters per second)? Choose 1 answer: Choose 1 answer: (Choice A) A $492$ (Choice B) B $-492$ (Choice C) C $348$ (Choice D) D $-348$ The volume of a square prism with base side $s$ and height $h$ is $s^2h$.
Setting up the math Let... $s(t)$ denote the base side of the prism at time $t$, $h(t)$ denote the height of the prism at time $t$, and $V(t)$ denote the volume of the prism at time $t$. We are given that $s'(t)=5$ and $h'(t)=-2$ (notice that $h'$ is negative). We are also given that that $s(t_0)=6$ and $h(t_0)=7$ for a specific time $t_0$. We want to find $V'(t_0)$. Relating the measures The measures relate to each other through the formula for the volume of a square prism: $V(t)=[s(t)]^2h(t)$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=2s(t)s'(t)h(t)+[s(t)]^2h'(t)$ Using the information to solve Let's plug ${s(t_0)}={6}$, ${s'(t_0)}={5}$, ${h(t_0)}={7}$, and $C{h'(t_0)}=C{-2}$ into the expression for $V'(t_0)$ : $\begin{aligned} V'(t_0)&=2{s(t_0)}{s'(t_0)}{h(t_0)}+[{s(t_0)}]^2C{h'(t_0)} \\\\ &=2({6})({5})({7})+({6})^2(C{-2}) \\\\ &=348 \end{aligned}$ In conclusion, the rate of change of the volume of the prism at that instant is $348$ cubic meters per second. Since the rate of change is positive, we know that the volume is increasing.